Thermo solution8 - Copy

Thermo solution8 - Copy - 650:351 Thermodynamics Doyle...

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Unformatted text preview: 650:351 Thermodynamics Doyle Knight Department of Mechanical and Aerospace Engineering Rutgers - The State University of New Jersey New Brunswick, NJ USA Tel: 732 445 4464 Email: doyleknight@gmail.com Office hours: Tuesday and Friday, 9:00 am to 11:30 am and by appointment Homework Assignment No. 8 Problem 9.35 A flow of 2 kg/s saturated vapor R-22 at 500 kPa is heated at constant pressure to 60 C. The heat is supplied by a heat pump that receives heat from the ambient at 300 K and work input. Assume the process is reversible and find the rate of work input. Solution The control volume is the heat exchanger. From the conservation of mass, m i = m e = m . From the First Law of Thermodynamics, 0 = Q H + mh i mh e since the change in kinetic and potential energy across the heat exchanger is negligible. From Table B.4.1, for saturated R-22 vapor, h g = 249 . 95 kJ/kg at p = 497 . 6 kPa and h g = 251 . 73 kJ/kg at p = 583 . 8 kPa 650:351 Thermodynamics Page 1 Homework Assignment No. 8 Problem 9.35 Thus, interpolating to p = 500 kPa, h 1 = 250 . 0 kJ/kg and s 1 = 0 . 9267 kJ/kg-K Note that T 1 = 0 . 14 C. Since T 2 = 60 C and p 2 = p 1 , the R-22 exits the heat exchanger as superheated vapor. From Table B.4.2, at p = 500 kPa and T = 60 C, h 2 = 293 . 22 kJ/kg and s 2 = 1 . 0696 kJ/kg-K Thus, Q H = m ( h 2 h 1 ) = (2)(293 . 22 250 . 0) = 86 . 44 kJ/s Consider a control volume enclosing the heat pump and heat exchanger. The First Law of Thermodynamics is 0 = Q L W + mh i mh e Why is Q H not included ? The Second Law of Thermodynamics is 0 = m i s i m e s e + Q L T L + S gen Why is Q H not included ? 650:351 Thermodynamics Page 2 Homework Assignment No. 8 Problem 9.35 Since the problem statement indicates that the process is reversible, S gen = 0. Thus, 0 = ms i ms e + Q L T L Thus, Q L = mT L ( s 2 s 1 ) = (2)(300)(1 . 0696 . 9267) = 85 . 74 kW From the First Law of Thermodynamics applied to the control volume enclosing the heat pump and heat exchanger, W = Q L + m ( h i h e ) = 85 . 74 + (2)(250 293 . 22) = . 7 kW Note that work is done on the control volume and hence W is negative. 650:351 Thermodynamics Page 3 Homework Assignment No. 8 Problem 9.36 A reversible steady-state device receives a flow of 1 kg/s air at 400 K and 450 kPa, and the air leaves at 600 K and 100 kPa. Heat transfer of 800 kW is added from a 1000 K reservoir, 100 kW is rejected at 350 K, and some heat transfer takes place at 500 K. Find the heat transferred at 500 K and the rate of work produced. Solution The control volume is the device. The conservation of mass is m i = m e = m The First Law of Thermodynamics is 0 = Q 1 Q 2 + Q 3 W + mh i mh e Note sign for Q 2 which is assumed positive for heat removed The Second Law of Thermodynamics is 0 = ms i ms e + summationdisplay Q T + S gen 650:351 Thermodynamics Page 4 Homework Assignment No. 8Homework Assignment No....
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This homework help was uploaded on 04/07/2008 for the course MECHANICAL 351 taught by Professor Knight during the Spring '08 term at Rutgers.

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Thermo solution8 - Copy - 650:351 Thermodynamics Doyle...

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